Answer:
1.76 * 6.02*10^23 = 1.05952*10^24
1.05952*2 = 2.11904 *10^24 oxygen and 1.05952*10^24 sulfur atoms
If it is assumed that there are only two isotopes then the percent abundance needs to add up to 100%
100-35= 65%
The second isotope will have a 65% abundance.
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The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 

Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)


Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.
Answer:
See explanation
Explanation:
If we look at the electron configuration closely, we will discover that the element must have had a ground state electron configuration of 2,4.
This is because, the innermost shell usually holds two electrons while the outer shells hold eight electrons each. The four electrons must be accommodated in the second shell in the ground state configuration of the compound.
However, when the atom is excited, one electron from this shell may move to the third shell to give the excited state configuration 2-3-1 as shown in the question.